I have kicking this idea around in my head for a couple weeks now. is it possible to have a driver stage that opperates in push-pull? Wouldn't it have to be Interstage coupled? I have no idea, just a guess

Hey-Hey!!!,
If the output is PP, then I don't see the need for an IT. If you want to drive a SE load, then I think you're going to have to use a TX. Seems that the extra complexity of a PP driver stage for a SE amp goes against the Holy Texts...

If I understand you we have had push pull driver stages for ever
The Williamson is a classic design with a push pull driver!
ITs a voltage amp followed by a concertina phase splitter attached to a low impedance cathode coupled driver stage (push-pull) that then feeds the output valves . This was done in 1947 i think and became the first commercial amplifier to deliver its rated power with less than 0.1 % distortion
regards Trev

Not really, IMHO. The Williamson uses a differential amplifier as its driver. If the two halves of the driver used separate cathode resistors, then it would be push-pull. You might want to do this if you wanted shunt negative feedback from the OP tube plates, either to the driver plates (to give the drivers a higher output impedance) or to the driver cathodes (to provide a voltage divider for the NFB). I personally think the second option would be a particularly good idea to improve the Williamson.

Yes I realise that the driver stage is a differential in the Williamson but the output is push pull!
Also It would not!!! be a good idea to change to seperate cathode resistors as this was a distortion cancelling trick that Williamson came up with . It is often referred to in design notes
Regards Trev

Originally posted by ray_moth
Not really, IMHO. The Williamson uses a differential amplifier as its driver. If the two halves of the driver used separate cathode resistors, then it would be push-pull.

Hey-Hey!!,
Ray, I'll have to disagree with you on this one. It is the output which sets the definition. Would you say an output stage is not PP if it has a common cathode load CCS? I certainly would not...

I do agree with the rest, the seperate cathode resistors of a PP driver stage( fed a balanced signal ) is an excellent way of delivering FB. Look at the 6V6 RCA amps, both PP and PPP for an ancient example. These amps used a concertina to create the first PP signal. 6AU6's with plate loads about 3x the size of the FB resistor from their cathode to the plate of the output tube.
cheers,
Douglas

Also It would not!!! be a good idea to change to seperate cathode resistors as this was a distortion cancelling trick that Williamson came up with . It is often referred to in design notes

Would so!!! Actually, I agree that simply separating the drivers' shared cathode resistor into two separate resistors, with no other change at all, would be a bad idea.

What I meant was that the second NFB option would be a good idea for the Williamson, i.e. balanced NFB loops from the OP tube plates to the driver cathodes. This would require separate driver cathode resistors to make it work.

As long as the Williamson's original triode-connected OP stage is used, adding more NFB loops is perhaps not so much of an issue (although I still think it's a good idea). However, if you wanted to use a pentode OP stage, or even UL, then you would require more NFB to get a decent damping factor. The design of the Williamson is not conducive to increasing NFB significantly using the existing global loop, because of potential instability due to having two coupling caps and the OP transformer inside the existing global NFB loop.

I believe that adding another, internal NFB loop, which would include the output tubes and drivers but exclude the input and splitter stages and the OP transformer, would enable NFB to be increased sufficiently to achieve adequate damping for pentode or UL output without incurring problems of instability. Williamson's design, with the splitter coming before the drivers, makes this approach feasible.